Question

The non-permissible value of x in the expression:

1/(sqrt(3) - sqrt(x))

Answer 1

Well there isn't any value that would give us 1/0, so that's not a problem. However, square roots are not defined for numbers less than 0. Because we have sqrt(x), x cannot be less than zero. So any negative number is non-permissible.

Answer 2

three

Answer 3

Oh yes, I failed to notice that one, my apologies.

Answer 4

square roots for number less than zero are complex numbers

Answer 5

It depends what field you are in. I will suppose you are working with the real numbers...
I think you have \[1/(\sqrt{3}-\sqrt{x})\]
You cannot divide by 0 (this is undefined.
That means you have to solve for when the denominator equals 0.
\[\sqrt{3}-\sqrt{x}=0\]
\[\sqrt{3}=\sqrt{x}\] Adding sqrt(x) to both sides
3=x (squaring both sides)
That means when x=3 we will have 1/0 and so
\[x \neq3\]
Also if we are working with the real numbers,
when we have the square root we cannot have what's inside of it be less than 0 (negative)

You have \[\sqrt{x}\]
so that means \[x \ge0\]

or said another way
x cannot be less than or equal to 0.

SOLUTION: x cannot equal 3 and x cannot be less than or equal to 0.

Answer 6

i dont see why complex numbers would be non-permissible

Answer 7

Thanks guys, I should of specified by the question says "Assuming x is positive"

Answer 8

IF YOU ARE IN THE FIELD: THE REAL NUMBERS
THEN THE COMPLEX NUMBERS WOULD NOT BE PERMISSIBLE.

That is why I said "I'll assume/suppose you are working with the real numbers"

Answer 9

@ Leaper: In that case my answer is applicable except for the part where we have x is greater than or equal to 0, you will now have x is greater than 0. So x cannot be 3 and x cannot be less than 0.
Recall that 0 is neither positive or negative.
(And i found a mistake above in my answer but after this quick explanation it is void)